src/Doc/Tutorial/Datatype/Nested.thy
author desharna
Fri, 27 May 2022 16:16:45 +0200
changeset 75466 5f2a1efd0560
parent 69597 ff784d5a5bfb
permissions -rw-r--r--
added predicate totalp_on and abbreviation totalp

(*<*)
theory Nested imports ABexpr begin
(*>*)

text\<open>
\index{datatypes!and nested recursion}%
So far, all datatypes had the property that on the right-hand side of their
definition they occurred only at the top-level: directly below a
constructor. Now we consider \emph{nested recursion}, where the recursive
datatype occurs nested in some other datatype (but not inside itself!).
Consider the following model of terms
where function symbols can be applied to a list of arguments:
\<close>
(*<*)hide_const Var(*>*)
datatype ('v,'f)"term" = Var 'v | App 'f "('v,'f)term list"

text\<open>\noindent
Note that we need to quote \<open>term\<close> on the left to avoid confusion with
the Isabelle command \isacommand{term}.
Parameter \<^typ>\<open>'v\<close> is the type of variables and \<^typ>\<open>'f\<close> the type of
function symbols.
A mathematical term like $f(x,g(y))$ becomes \<^term>\<open>App f [Var x, App g
  [Var y]]\<close>, where \<^term>\<open>f\<close>, \<^term>\<open>g\<close>, \<^term>\<open>x\<close>, \<^term>\<open>y\<close> are
suitable values, e.g.\ numbers or strings.

What complicates the definition of \<open>term\<close> is the nested occurrence of
\<open>term\<close> inside \<open>list\<close> on the right-hand side. In principle,
nested recursion can be eliminated in favour of mutual recursion by unfolding
the offending datatypes, here \<open>list\<close>. The result for \<open>term\<close>
would be something like
\medskip

\input{unfoldnested.tex}
\medskip

\noindent
Although we do not recommend this unfolding to the user, it shows how to
simulate nested recursion by mutual recursion.
Now we return to the initial definition of \<open>term\<close> using
nested recursion.

Let us define a substitution function on terms. Because terms involve term
lists, we need to define two substitution functions simultaneously:
\<close>

primrec
subst :: "('v\<Rightarrow>('v,'f)term) \<Rightarrow> ('v,'f)term      \<Rightarrow> ('v,'f)term" and
substs:: "('v\<Rightarrow>('v,'f)term) \<Rightarrow> ('v,'f)term list \<Rightarrow> ('v,'f)term list"
where
"subst s (Var x) = s x" |
  subst_App:
"subst s (App f ts) = App f (substs s ts)" |

"substs s [] = []" |
"substs s (t # ts) = subst s t # substs s ts"

text\<open>\noindent
Individual equations in a \commdx{primrec} definition may be
named as shown for @{thm[source]subst_App}.
The significance of this device will become apparent below.

Similarly, when proving a statement about terms inductively, we need
to prove a related statement about term lists simultaneously. For example,
the fact that the identity substitution does not change a term needs to be
strengthened and proved as follows:
\<close>

lemma subst_id(*<*)(*referred to from ABexpr*)(*>*): "subst  Var t  = (t ::('v,'f)term)  \<and>
                  substs Var ts = (ts::('v,'f)term list)"
apply(induct_tac t and ts rule: subst.induct substs.induct, simp_all)
done

text\<open>\noindent
Note that \<^term>\<open>Var\<close> is the identity substitution because by definition it
leaves variables unchanged: \<^prop>\<open>subst Var (Var x) = Var x\<close>. Note also
that the type annotations are necessary because otherwise there is nothing in
the goal to enforce that both halves of the goal talk about the same type
parameters \<open>('v,'f)\<close>. As a result, induction would fail
because the two halves of the goal would be unrelated.

\begin{exercise}
The fact that substitution distributes over composition can be expressed
roughly as follows:
@{text[display]"subst (f \<circ> g) t = subst f (subst g t)"}
Correct this statement (you will find that it does not type-check),
strengthen it, and prove it. (Note: \<open>\<circ>\<close> is function composition;
its definition is found in theorem @{thm[source]o_def}).
\end{exercise}
\begin{exercise}\label{ex:trev-trev}
  Define a function \<^term>\<open>trev\<close> of type \<^typ>\<open>('v,'f)term => ('v,'f)term\<close>
that recursively reverses the order of arguments of all function symbols in a
  term. Prove that \<^prop>\<open>trev(trev t) = t\<close>.
\end{exercise}

The experienced functional programmer may feel that our definition of
\<^term>\<open>subst\<close> is too complicated in that \<^const>\<open>substs\<close> is
unnecessary. The \<^term>\<open>App\<close>-case can be defined directly as
@{term[display]"subst s (App f ts) = App f (map (subst s) ts)"}
where \<^term>\<open>map\<close> is the standard list function such that
\<open>map f [x1,...,xn] = [f x1,...,f xn]\<close>. This is true, but Isabelle
insists on the conjunctive format. Fortunately, we can easily \emph{prove}
that the suggested equation holds:
\<close>
(*<*)
(* Exercise 1: *)
lemma "subst  ((subst f) \<circ> g) t  = subst  f (subst g t) \<and>
       substs ((subst f) \<circ> g) ts = substs f (substs g ts)"
apply (induct_tac t and ts rule: subst.induct substs.induct)
apply (simp_all)
done

(* Exercise 2: *)

primrec trev :: "('v,'f) term \<Rightarrow> ('v,'f) term"
  and trevs:: "('v,'f) term list \<Rightarrow> ('v,'f) term list"
where
  "trev (Var v)    = Var v"
| "trev (App f ts) = App f (trevs ts)"
| "trevs [] = []"
| "trevs (t#ts) = (trevs ts) @ [(trev t)]" 

lemma [simp]: "\<forall> ys. trevs (xs @ ys) = (trevs ys) @ (trevs xs)" 
apply (induct_tac xs, auto)
done

lemma "trev (trev t) = (t::('v,'f)term) \<and> 
       trevs (trevs ts) = (ts::('v,'f)term list)"
apply (induct_tac t and ts rule: trev.induct trevs.induct, simp_all)
done
(*>*)

lemma [simp]: "subst s (App f ts) = App f (map (subst s) ts)"
apply(induct_tac ts, simp_all)
done

text\<open>\noindent
What is more, we can now disable the old defining equation as a
simplification rule:
\<close>

declare subst_App [simp del]

text\<open>\noindent The advantage is that now we have replaced \<^const>\<open>substs\<close> by \<^const>\<open>map\<close>, we can profit from the large number of
pre-proved lemmas about \<^const>\<open>map\<close>.  Unfortunately, inductive proofs
about type \<open>term\<close> are still awkward because they expect a
conjunction. One could derive a new induction principle as well (see
\S\ref{sec:derive-ind}), but simpler is to stop using
\isacommand{primrec} and to define functions with \isacommand{fun}
instead.  Simple uses of \isacommand{fun} are described in
\S\ref{sec:fun} below.  Advanced applications, including functions
over nested datatypes like \<open>term\<close>, are discussed in a
separate tutorial~@{cite "isabelle-function"}.

Of course, you may also combine mutual and nested recursion of datatypes. For example,
constructor \<open>Sum\<close> in \S\ref{sec:datatype-mut-rec} could take a list of
expressions as its argument: \<open>Sum\<close>~@{typ[quotes]"'a aexp list"}.
\<close>
(*<*)end(*>*)